Problem: The lifespans of bears in a particular zoo are normally distributed. The average bear lives $45.2$ years; the standard deviation is $10.6$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a bear living less than $24$ years.
Solution: $45.2$ $34.6$ $55.8$ $24$ $66.4$ $13.4$ $77$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $45.2$ years. We know the standard deviation is $10.6$ years, so one standard deviation below the mean is $34.6$ years and one standard deviation above the mean is $55.8$ years. Two standard deviations below the mean is $24$ years and two standard deviations above the mean is $66.4$ years. Three standard deviations below the mean is $13.4$ years and three standard deviations above the mean is $77$ years. We are interested in the probability of a bear living less than $24$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the bears will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the bears will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $24$ years and the other half $({2.5\%})$ will live longer than $66.4$ years. The probability of a particular bear living less than $24$ years is ${2.5\%}$.